Answer K = 200
h = (.005)(.65) = .00325/unit
µ h = 3.25/thousand.
r = (22, 34, 32, 12, 8, 44, 54, 16, 76, 30)
a) Silver Meal Solution (S/M)
Start in period 1:
C(1) = 200
C(2) = [200 + ((3.25)(34)]/2 = 172.8
C(3) = [(155.25)(2) + (3.25)(2)(32)]/3 = 183.5 Stop.
y1 = r1 + r2 = 56
Start in period 3:
C(1) = 200
C(2) = [200 + (3.25)(12)]/2 = 119.5
C(3) = [(2)(119.5) + (2)(3.25)(8)]/3 = 97
C(4) = [(3)(97) + (3)(3.25)(44)]/4 = 180 Stop.
y3 = r3 + r4 + r5 = 32 + 12 + 8 = 52
Start in period 6:
C(1) = 200
C(2) = [200 + (3.25)(54)]/2 = 187.75
C(3) = [(2)(187.75) + (2)(3.25)(16)]/3 = 159.83...
C(4) = [(3)(159.833..) + (3)(3.25)(76)]/4 = 305.125 Stop.
y6 = r6 + r7 + r8 = 44 + 54 + 16 = 114
Obtain y9 = r9 + r10 = 106
Silver Meal Solution = (56, 0, 52, 0, 0, 114, 0, 0, 106, 0)
Cost = (4)(200) + (3.25)(34 + 20 + 8 + 70 + 16 + 30)
= $1378.50
Answer C:- Least Unit Cost (LUC) method:
Start in period 1:
C(1) = 200/22 = 9.09
C(2) = [200 + (3.25)(34)]/(22 + 34) = 5.54
C(3) = [200+(3.25)(34+(2)(32))]/(22+34+32) = 5.89 Stop.
y1 = r1 + r2 = 56
Start in period 3:
C(1) = (200)/(32) = 6.25
C(2) = [200 + (3.25)(12)]/(32 + 12) = 5.43
C(3) = [200 + (3.25)(12+(2)(8))]/(32+12 8) = 5.60 Stop.
y3 = r3 + r4 = 44
Start in period 5:
C(1) = 200/8 = 25
C(2) = [200+(3.25)(44)]/(8 + 44) = 6.60
C(3) = [200+(3.25)(44+(54)(2))]/(8+44+54) = 6.55
C(4) = [200+(3.25)(44+(54)(2)+(16)(3))]/(8+44+54+16)= 7.97 Stop.
y5 = r5 + r6 + r7 = 106
Start in period 8:
C(1) = 12.5
C(2) = 4.9
C(3) = 5.3 Stop
y8 = r8 + r9 = 92
y10 = r10 = 30
Answer d-
Summary of Results
Cost
S/M $1378.50
LUC $1890.50
*Note: None of these solutions is optimal. The true optimal solution is (56, 0, 52, 0, 0, 44, 70, 0, 106, 0) at a cost of $1351.00. Silver Meal obviously comes the closest for this example.
LUC solution: (56, 0, 44, 0, 106, 0, 0, 92, 0, 30)
Cost = (200)(5)+(3.25)(34 + 12 + 98 + 54 + 76) = $1,890.50.
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